Question

Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{4}{3}$ is applied to triangle $XYZ,$ to obtain triangle $X'Y'Z'.$ Let $A$ be the area of the region that is contained in both triangles $XYZ$ and $X'Y'Z'.$ Find $\frac{A}{[XYZ]}.$

Answer 1

Explanation:

Let each side of triangle XYZ have length s. We therefore have:

s = YZ = YQ + QR + RZ = x + 2y

Since each side of triangle X1 Y2 Z2 is 3/4 the length of the corresponding side of triangle XYZ, we have:

3/4s = X1 Z1 = X1 Q + QP + P Z1 = 2x +y.

The system of equations:

x + 2y = s

2x + y = 3/4s

Multiply the second equation by 2, then subtract the first equation.

3x = 1/2s, so x = 1/6s.

Now we have y = 3/4s-2x = 5/12s.

A/[XYZ] = [XYZ] - 3[YPQ]/[XYZ]

= {[XYZ] - 3[(25)/144] * [XYZ]} / [XYZ]

={23/48* [XYZ]} / [XYZ]

= 23/48